A firefighting crew uses a water cannon that shoots water at 25.0m/s at a fixed angle of 53.0 degrees above the horizontal. The firefighters want to direct the water at a blaze that is 10.0 m above ground level.
How far from the building should they position their cannon? There are two possibilities.

Explanation:

Consider the vertical motion of water,

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 25sin53 = 19.97 m/s

        Acceleration, a = -9.81 m/s²  

        Displacement, s =10 m      

     Substituting

                      s = ut + 0.5 at²

                      10 = 19.97 x t + 0.5 x -9.81 xt²

                      t = 3.49 s  or  t = 0.59 s

     So at times 3.49 s and 0.59 s water reaches at a height 10 m

Now consider the horizontal motion of water,

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 25cos53 = 15.05 m/s

        Acceleration, a = 0 m/s²  

        Displacement, s =?

         time, t = 3.49 s or 0.59 s

     Substituting

                      s = ut + 0.5 at²

                      s = 15.05 x 3.49 + 0.5 x 0 x3.49²

                      s = 52.52 m

                      s = 15.05 x 0.59 + 0.5 x 0 x0.59²

                      s = 8.88 m

They should position cannon at a distance of 8.88 m or 52.52 m from building.

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